题意：

给出一个存钱罐的重量和没存钱之前存钱罐的重量，然后给出几种硬币的重量和币值，计算存钱罐里至少有多少钱。

题目：

Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.

But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it’s weight in grams.

Output

Print exactly one line of output for each test case. The line must contain the sentence “The minimum amount of money in the piggy-bank is X.” where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line “This is impossible.”.

Sample Input

3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4

Sample Output

The minimum amount of money in the piggy-bank is 60.
The minimum amount of money in the piggy-bank is 100.
This is impossible.

分析：

（1）、关于恰好装满问题
是否恰好装满的解法不同只在于初始值的不同
恰好装满：1、求最大值时，除了dp[0] 为0，其他都初始化为无穷小 -inf;2、求最小值时，除了dp[0] 为0，其他都初始化为无穷大inf(求最小值时原状态为inf【状态为不存在此种情况】+一个数一定大于inf，此时选择小值为inf，状态不变);
不必恰好装满： 全初始化为0.
(2)首先满的猪的重量减去空的猪的重量这样就可以获得能装多少硬币的重量。然后用完全背包，看是否能恰好装满且价值最小。
这里推荐一篇写的好的博文写挺好的，就是太监了的一篇博客，狗头保命

AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int M=5e2+10;
int t,a,b,n;
int x[M],y[M],dp[10010];
int main()
{scanf("%d",&t);while(t--){memset(dp,inf,sizeof(dp));scanf("%d%d%d",&a,&b,&n);dp[0]=0;//恰好装满，只有背包容量为0时满足，其他初始化为无穷表示未被定义int ma=b-a;for(int i=1; i<=n; i++)scanf("%d%d",&x[i],&y[i]);for(int i=1; i<=n; i++)for(int j=y[i]; j<=ma; j++)//正序，放入第i物品,之前的状态需要进行改变,故需要正序dp[j]=min(dp[j],dp[j-y[i]]+x[i]);if(dp[ma]==inf)printf("This is impossible.\n");elseprintf("The minimum amount of money in the piggy-bank is %d.\n",dp[ma]);}return 0;
}